For the reaction Cu^{2+}_{(aq)} + Fe_{(s)} ri | vedclass

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(D) The cell reaction is:Anode: $Fe_{(s)} 	o Fe^{2+}_{(aq)} + 2e^-$Cathode: $Cu^{2+}_{(aq)} + 2e^- 	o Cu_{(s)}$Number of electrons transferred,$n =

Vedclass · Accepted Answer

$-150540$

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(D) The cell reaction is:Anode: $Fe_{(s)} 	o Fe^{2+}_{(aq)} + 2e^-$Cathode: $Cu^{2+}_{(aq)} + 2e^- 	o Cu_{(s)}$Number of electrons transferred,$n = 2$.Standard cell potential: $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{Cu^{2+}/Cu} - E^o_{Fe^{2+}/Fe} = 0.34 \ V - (-0.44 \ V) = +0.78 \ V$.Using the formula: $\Delta G^o = -nFE^o_{cell}$.Given $F = 96500 \ C \ mol^{-1}$.$\Delta G^o = -(2) 	imes (96500 \ C \ mol^{-1}) 	imes (0.78 \ V) = -150540 \ J \ mol^{-1}$.Thus,the correct option is $D$.

For the reaction $Cu^{2+}_{(aq)} + Fe_{(s)} \rightleftharpoons Fe^{2+}_{(aq)} + Cu_{(s)}$,calculate $\Delta G^o$ in Joules.
Given: $E^o_{Cu^{2+}/Cu} = +0.34 \ V$,$E^o_{Fe^{2+}/Fe} = -0.44 \ V$.

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For the reaction $Cu^{2+}_{(aq)} + Fe_{(s)} \rightleftharpoons Fe^{2+}_{(aq)} + Cu_{(s)}$,calculate $\Delta G^o$ in Joules.Given: $E^o_{Cu^{2+}/Cu} = +0.34 \ V$,$E^o_{Fe^{2+}/Fe} = -0.44 \ V$.